A box contains $3$ red jelly beans, $6$ green jelly beans, and $2$ blue jelly beans. If we choose a jelly bean, then another jelly bean without putting the first one back in the box, what is the probability that the first jelly bean will be red and the second will be red as well? Write your answer as a simplified fraction.
Solution: The probability of event A happening, then event B, is the probability of event A happening times the probability of event B happening given that event A already happened In this case, event A is picking a red jelly bean and leaving it out. Event B is picking another red jelly bean. Let's take the events one at at time. What is the probability that the first jelly bean chosen will be red? There are $3$ red jelly beans, and $11$ total, so the probability we will pick a red jelly bean is $\dfrac{3} {11}$ After we take out the first jelly bean, we don't put it back in, so there are only $10$ jelly beans left. Also, we've taken out one of the red jelly beans, so there are only $2$ left altogether. So, the probability of picking another red jelly bean after taking out a red jelly bean is $\dfrac{2} {10}$ Therefore, the probability of picking a red jelly bean, then another red jelly bean is $\dfrac{3}{11} \cdot \dfrac{2}{10} = \dfrac{3}{55}$